Question

Two boxes are connected by a rope strung over a pulley. Box A with a mass of 5.0 kg sits on the surface of a plane inclined with an angle of 33 degrees. The coefficient of static and kinetic friction between the box A and the surface it sits on is 0.32 and 0.20 respectively. The rope connecting Box A and Box B is looped over a frictionless pulley at the top of the incline and Box B hangs free in the air suspended below the pulley. If both boxes are motionless, what is the mass of Box B? If Box B is one kilogram heavier than that what is the acceleration of the boxes?

Answer #1

s = coeffcient of static friction = 0.32

k = coeffcient of kinetic friction = 0.20

M_{a} = mass of box A = 5 kg

M_{b} = mass of box B

**for mass A :**

force equation perpendicular to incline plane is given as

**F _{n} = M_{a} g Cos**

**F _{n} =** 5 x 9.8 Cos33 =

**static friction force acting**

**f _{s} =** s

**kinetic friction force**

**f _{k} =** k

force equation parallel to incline plane is given as

T = M_{a} g Sin** +
f _{s}**

T = 5 x 9.8 Sin33 + 13.2

T = 39.88 N

**for mass B :**

force equation is given as

**T = M _{b} g**

39.88 = M_{b} (9.8)

**M _{b} = 4.1 kg**

ii)

new mass of Block B = **M' _{b} = 4.1 + 1 = 5.1
kg**

**weight of box B = 5.1 x 9.8 = 49.98 N**

force equation for the motion of block B

**M _{b}g - T = M_{b}a**

5.1 x 9.8 - T = 5.1a

**49.98 - T = 5.1
a **

**T = 49.98 - 5.1 a eq-1**

force equation for the motion of block A

**T - M _{a} g Sin - f_{k}
= M_{a} a**

(49.98 - 5.1 a) - 5 x 9.8 Sin33 - 8.22 = 5a

**a = 1.5 m/s ^{2}**

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